The solution of the differential equation `(dy)/(dx)=(x+y)/(x)` satisfying the conditions y(1)=1 is

Options

A. `y=x log x +x^(2)`
B. `y= x e^((x-1))`
C. `y= x log x +x`
D. `y=log x +x`

Answer

To solve the differential equation \(\frac{dy}{dx} = \frac{x+y}{x}\) with the initial condition \(y(1) = 1\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given differential equation:

\[ \frac{dy}{dx} = \frac{x + y}{x} \]

This can be rewritten as:

\[ \frac{dy}{dx} = 1 + \frac{y}{x} \]

### Step 2: Rearrange the equation
Next, we rearrange the equation to isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} - \frac{y}{x} = 1 \]

This is now in the standard form of a linear differential equation:

\[ \frac{dy}{dx} + P(x)y = Q(x) \]

where \(P(x) = -\frac{1}{x}\) and \(Q(x) = 1\).

### Step 3: Find the integrating factor
The integrating factor \(I(x)\) is given by:

\[ I(x) = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln |x|} = \frac{1}{x} \]

### Step 4: Multiply through by the integrating factor
We multiply the entire differential equation by the integrating factor:

\[ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = \frac{1}{x} \]

This simplifies to:

\[ \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{1}{x} \]

### Step 5: Integrate both sides
Now, we integrate both sides:

\[ \int \frac{d}{dx}\left(\frac{y}{x}\right) \, dx = \int \frac{1}{x} \, dx \]

This gives us:

\[ \frac{y}{x} = \ln |x| + C \]

### Step 6: Solve for \(y\)
Multiplying through by \(x\) to solve for \(y\):

\[ y = x \ln |x| + Cx \]

### Step 7: Apply the initial condition
Now we apply the initial condition \(y(1) = 1\):

\[ 1 = 1 \cdot \ln(1) + C \cdot 1 \]

Since \(\ln(1) = 0\), we have:

\[ 1 = C \]

### Step 8: Write the final solution
Substituting \(C = 1\) back into the equation for \(y\):

\[ y = x \ln |x| + x \]

Thus, the solution to the differential equation satisfying the condition \(y(1) = 1\) is:

\[ \boxed{y = x \ln x + x} \]

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